I am trying to read Hoffman Kunze's book on linear algebra and I have a doubt in a particular result, (Theorem 1) of Section 5.2. Specifically, the theorem states:
Let $n > 1$ and let $D$ be an alternating $(n - 1)$-linear function on $(n - 1)\times (n - 1)$ matrices over $K$. For each $j$, $1 < j \le n$, the function $E_j$ defined by $$E_j(A) = \sum_{i=1}^n(-l)^{i+j}A_{ij}D_{ij}$$ is an alternating $n$-linear function on $n \times n$ matrices $A$. If $D$ is a determinant function, so is each $E_j$.
Here $D_{ij}=D[A(i|j)]$ where $A(i|j)$ denotes the matrix obtained by deleting the $i$th row and the $j$th column of $A$.
Now my question concerns the $n$-linear part. I understand why $D_{ij}$ is linear in every row except the $i$th row and that $D_{ij}$ is independent of the $i$th row. What I do not understand is why $D_{ij}$ is linear in the $i$th row.
For example, if $n=2$ and $D([a])=a$ then $$D_{11}\begin{pmatrix} a+a'& b+b'\\c & d\end{pmatrix}=d$$ while $$D_{11}\begin{pmatrix} a& b\\c & d\end{pmatrix}+D_{11}\begin{pmatrix} a'& b'\\c & d\end{pmatrix}=d+d=2d.$$
Yet the authors state $A_{ij}D_{ij}$ is $n$-linear.
As you have observed, $D_{ij}(A)$ is linear in every row except the $i$th row, and $D_{ij}(A)$ is independent of row $i$. On the other hand, $A_{ij}$ is independent in every row except the $i$th row, and $A_{ij}$ is linear in row $i$. Thus, $A_{ij}D_{ij}(A)$ is linear in every row!